# Triangle congruence when the longest sides, the largest angles, and one of the other sides are congruent?

Are two scalene triangles (a scalene triangle is one with no two equal sides) congruent if their longest sides, largest angles, and shortest sides are congruent?

I believe the case described above is a special case of side-side-angle that actually implies congruence of the two triangles involved. Please let me know if you can come up with any counterexample, that is, two triangles that are not congruent and satisfy the conditions described above. I couldn't come up with such an example.

Here is a counterexample, not for the problem above, but for the general side-side-angle case where shows side-side-angle in general is not a sufficient condition for congruence of two triangles.

Here is a close answer on StackExchange, but it still doesn't answer my question.

Here is my attempt to solve the problem---side-side-angle special case

We want to prove that given the largest and the smallest sides of a triangle along with the measure of its largest angle, we can construct one and only one such triangle.

The largest angle ought to be opposite the longest side. Thus, the shortest side shall be adjacent to the largest angle.

We start to construct the triangle by drawing the shortest side AB. Then we draw the ray AC such that the angle BAC becomes the largest angle in the triangle. Note that the angle BAC can be a right, obtuse, or acute angle without violating the assumption that the side BC is the largest side (See figure parts a, b, and c).

(an interesting but irrelevant question may be "What is the smallest possible measure of the largest angle in an scalene triangle given the size of its largest and smallest sides?"---can be solved using the sine rule)

Now we take the last step in constructing the triangle by drawing an arc centered at B with a radius $$r$$ which equals the length of the longest side BC. Then, we claim that we can uniquely find the third vertex C of the triangle where the arc intersects the ray AC.

The arc shall not intersect the ray AC in more than one point because that would require the arc to intersect the side AB as well which implies that the side AB is longer than the side BC which contradicts our assumption. Thus, the point C is unique.

As an argument for the uniqueness of C, we can use sine rule. With your notation, $$\frac{\sin|\angle ACB|}{|AB|}=\frac{\sin|\angle BAC|}{BC}$$ As $$AB$$ is the shortest side, it is opposite the smallest angle. In particular, we must have that $$ACB$$ is acute, so sine rule uniquely determines it. So we can then determine all the angles, and all the sides.
Your construction can produce two choices for C if $$|\angle BAC|\leq \frac{\pi}{2}$$. But one of the points will give $$|AB|\geq |AC|$$, so won't give the prescribed triangle.
• @Adrian why $\angle ACB$ must be acute? Mar 24, 2020 at 21:13
• $\angle ACB$ is opposite the shortest side $AB$, so it is the smallest angle. If it were greater than 90 degrees, then $\angle ABC\geq 90$ and $\angle BAC \geq 90$, so $180=\angle ACB + \angle ABC + \angle BAC \geq 270$, which is a contradiction. Mar 24, 2020 at 21:21